BZOJ4378: [POI2015]Logistyka

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每个数对答案的贡献=min(a[i],s)

所以用树状数组维护小于等于每个数的权值和以及个数

每次对于一个s，把小于s的所有数加起来，再把大于s的当成s，统计一下贡献，看能否实现

#include<iostream>#include<cstdio>#include<map>#include<algorithm>#define N 2000010using namespace std;struct ppp {long long siz,sum;};ppp c[N];ppp operator+(ppp x,ppp y){return (ppp){x.siz+y.siz,x.sum+y.sum};}long long n,m;long long lowbit(long long x) {return x&(-x);}void change(long long x,ppp v){for(;x<=m+1;x+=lowbit(x))c[x]=c[x]+v;}ppp check(long long x){ppp ret={0,0};for(;x;x-=lowbit(x))ret=ret+c[x];return ret;}long long o[N],x[N],y[N];long long aa[N];map<long long,long long>p;long long a[N];int main(){scanf("%lld%lld",&n,&m);long long i,j;char s[10];for(i=1;i<=m;i++){scanf("%s%lld%lld",s,&x[i],&y[i]);if(s[0]=='Z') o[i]=1;aa[i]=y[i];}sort(aa+1,aa+m+1);p[0]=1;for(i=1;i<=m;i++)p[aa[i]]=i+1;change(p[0],(ppp){n,0});for(i=1;i<=m;i++){if(o[i]==0){change(p[a[x[i]]],(ppp){-1,-a[x[i]]});a[x[i]]=y[i];change(p[a[x[i]]],(ppp){1,a[x[i]]});}else{ppp t=check(p[y[i]]);//cout<<t.siz<<' '<<t.sum<<endl;if((n-t.siz)*y[i]+t.sum>=y[i]*x[i]) puts("TAK");else puts("NIE");}}}

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